Answer: Thus percentage purity of sodium carbonate is 60.83%.
Explanation:
[tex]Na_2CO_3 + 2HCl \longrightarrow 2NaCl + CO_2 + H_2O[/tex]
Volume of the HCl solution ,V = 15.55 mL= 0.01555 L
Concentration of HCl solution ,C= 0.1755 M
Moles of HCl in 0.1755 M solution = n
[tex]Molarity=\frac{n}{V(L)}[/tex]
[tex]n=C\time V=0.1755 M\times 0.01555 L = 2.729\times 10^{-3} mol[/tex]
According o reaction 2 moles of HCl reacts with 1 mole of sodium carbonate.
Then n moles of HCl will react with:
[tex]\frac{1}{2}\times n=\frac{1}{2}\times 2.729\times 10^{-3} mol=1.3645\times 10^{-3} mol[/tex] of sodium carbonate.
Mass of [tex]1.364\times 10^{-3} mol[/tex] of sodium carbonate:
[tex]1.364\times 10^{-3} mol\times 106 g/mol=0.144584 g[/tex]
Mass of the sample given= 0.2377 grams
Mass used up in reaction = 0.144585 g
percentage purity= [tex]\frac{0.144585}{0.2337}\times 100=60.83\%[/tex]
Thus percentage purity of sodium carbonate is 60.83%.
